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We consider the function $f(x,y)=x^2+y^2$ in $\omega = (0,1)^2.$ I am wondering about the existence of a $C^2-$extension $F$ of $f$ in $\Omega = (0,2)^2$ such that $F$ is harmonic in $\Omega-\overline{\omega}$. Thanks

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your $f$ is not harmonic – Will Jagy Aug 20 '12 at 1:42
sorry, it is not harmonic – med Aug 20 '12 at 15:47

$C^2$ is clearly too much to ask for, since it would imply the continuity of $\Delta F$. You have $\Delta F\equiv 1$ on $\omega$, and want $\Delta F\equiv 0$ on $\Omega\setminus\overline{\omega}$.

But even $C^1$ is impossible. Indeed, suppose $F$ is a $C^1$ extension with the desired property and let $G=F_z=\frac12(F_x-iF_y)$. The function $G$ is holomorphic in $\Omega\setminus\overline{\omega}$. In $\omega$ we have $G=(z\bar z)_z=\bar z$ (complex derivatives are a great time-saver). On the top boundary of $\omega$ we have $\bar z=z-2i$. By the boundary uniqueness, $G(z)\equiv z-2i$ in $\Omega\setminus \omega$. But this is inconsistent with the values on the right side of $\partial\omega$, where $\bar z\ne z-2i$.

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A minor nit: isn't the Laplacian of $F$ actually $4$ on $\omega$ ? – hardmath Aug 20 '12 at 4:01
@hardmath Yes if your Laplacian is $\dfrac{\partial ^2}{\partial x^2}+\dfrac{\partial ^2}{\partial y^2}$. But my Laplacian is $\dfrac{\partial^2}{\partial z \partial \bar z}$. Probabilist's Laplacian falls in between. – user31373 Aug 20 '12 at 4:24

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